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Hunter-Best [27]
3 years ago
15

The velocity component with which a projectile covers certain horizontal distance is maximum at the moment of? a) Hitting the gr

ound b) Highest point c) Projection d) None of these
Physics
1 answer:
Crazy boy [7]3 years ago
3 0
A) hitting the ground
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A box of unknown mass is sliding with an initial speed vi = 5.60 m/s across a horizontal frictionless warehouse floor when it en
Maksim231197 [3]

We know that the Delta E + W(Work done by non-conservative forces) = 0 (change of energy)

In here, the non-conservative force is the friction force where f = uN (u =kinetic friction coefficient) 

W= f x d = uNd ; N=mg 
Delta E = 1/2 mV^2 -1/2mVi^2 

umgd + 1/2mV^2 - 1/2mVi^2 = 0 (cancel out the m term) 

This will then give us: 

1/2Vi^2-ugd = 1/2V^2 

V^2 = Vi^2 - 2ugd

So plugging in our values, will give us:

V= Sqrt (5.6^2 -2.3^2)

=sqrt (26.07)

= 5.11 m/s 

 

6 0
3 years ago
7. What does the slope of a position-time graph represent?
allochka39001 [22]

Answer:

a. Velocity

Explanation:

The slope of the tangent line on a position-time graph is the instantaneous velocity.

7 0
3 years ago
A bullet with a mass 2.25g is fired up into the air with a velocity of 187.5 m/s. What is the maximum height of the bullet
Ksivusya [100]

Answer:

1793.7m

Explanation:

From the principle of conservation of energy; the kinetic energy substended by the object equals the potential energy sustain by the object when it gets to its maximum position.

Now the kinetic energy; is

K.E = 1/2 × m × v2

Where m is mass

v is velocity

Hence.

K.E = 1/2 × 2.25 × (187.5)^2

Now this should be same with the potential energy which is given as;

P.E = m× g× h

Where m is mass of object

g is acceleration of free fall due to gravity = 9.8m/S2

h is maximum height substain by the object.

Hence P.E = 2.25 × 9.8 × h

From the foregoing analysis of energy conversation it implies;

1/2 × 2.25 × (187.5)^2 =2.25 × 9.8 × h

=> 1/2 × (187.5)^2 = 9.8 × h

=>1/2 × (187.5)^2 / 9.8 = h

=> 1793.69m = h

h= 1793.69m

h =1793.7m to 1 decimal place

3 0
3 years ago
A celebrated Mark Twain story has motivated contestants in the Calaveras County Jumping Frog Jubilee, where frog jumps as long a
IrinaK [193]

The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

To find the answer, we need to know about the time of flight and range of projectile motion.

<h3>What's the expression of range of a projectile motion?</h3>
  • Range = U²× sin(2θ)/g
  • U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
  • U=√{Range×g/sin(2θ)}
  • Here, range= 2.20m, = 36.5°
  • U= √{2.20×9.8/sin(73)}

U= √{2.20×9.8/sin(73)} = 22.5m/s

<h3>What's the expression of time of flight in projectile motion?</h3>
  • Time of flight= (2×U×sinθ)/g
  • So, T= (2×22.5×sin36.5°)/9.8

= 2.73 s

Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

Learn more about the range and time period of projectile motion here:

brainly.com/question/24136952

#SPJ1

4 0
2 years ago
If the statement is true, select True. If it is false, select False.
RUDIKE [14]

Answer:

false

Explanation:

6 0
3 years ago
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